The scalar field $f(x, y) = x^2e^y - y^3$ has a critical point at $(0, 0)$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D) D The test is inconclusive
The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &=2xe^y \\ \\ f_y &= x^2e^y - 3y^2 \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= 2e^y = 2 \\ \\ f_{yx} &=2xe^y = 0 \\ \\ f_{xy} &= 2xe^y = 0 \\ \\ f_{yy} &= x^2e^y - 6y = 0 \end{aligned}$ Therefore, $H = (2)(0) - (0)(0) = 0$. Because $H$ is zero, the test is inconclusive.